Dr Ashok Kumar Probability

1

Dr. Ashok Kumar (Assistant Professor)

Department of MathematicsHNB Garhwal University Srinagar,

(UK)

Introduction to Probability TheoryIntroduction to Probability Theory

Basic Definitions: Events, Sample Space, and Probabilities

Basic Rules for ProbabilityConditional ProbabilityIndependence of EventsCombinatorial ConceptsThe Law of Total Probability and Baye’s Theorem

ProbabilityProbability

2-2

DefinitionsProbability experiment: An action through which

specific results (counts, measurements, or responses) are obtained.

Outcome: The result of a single trial in a probability experiment.

Sample Space: The set of all possible outcomes of a probability experiment

Event: One or more outcomes and is a subset of the sample space

Pattern Classification, Chapter 1

4

Sample SpaceThe sample space is the set of all possible outcomes.

Simple EventsThe individual outcomes are called simple events.

EventAn event is any collectionof one or more simple events

Events & Sample Spaces

HINTS TO REMEMBER:Probability experiment: Roll a six-sided

die

Sample space: {1, 2, 3, 4, 5 ,6}

Event: Roll an even number (2, 4, 6)

Outcome: Roll a 2, {2}

Identifying Sample Space of a Probability Experiment

A probability experiment consists of tossing a coin and then rolling a six-sided die. Describe the sample space.

There are two possible outcomes when tossing a coin—heads or tails. For each of these there are six possible outcomes when rolling a die: 1, 2, 3, 4, 5, and 6. One way to list outcomes for actions occurring in a sequence is to use a tree diagram. From this, you can see the sample space has 12 outcomes.

Tree Diagram for Coin and Die Experiment

{H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}

|

H1 T6T5T4T3T2T1H6H5H4H3H2

1

654321654321

HT

Some moreA probability experiment consists of

recording a response to the survey statement below and tossing a coin. Identify the sample space.

Survey: There should be a limit to the number of terms a US senator can serve.

Agree Disagree No opinion



A. Start a tree diagram by forming a branch for each possible response to the survey. Agree Disagree No opinion

Some moreA. probability experiment consists of


B. At the end of each survey response branch, draw a new branch for each possible coin outcome.Agree Disagree No opinion

H H HT T T



D. List the sample space{Ah, At, Dh, Dt, Nh, Nt}

Agree Disagree No opinion

H H HT T T

3. Rolling a two balanced dice – 36 outcomes

Set - a collection of elements or objects of interest Empty set (denoted by ∅)

a set containing no elements Universal set (denoted by S)

a set containing all possible elements Complement (Not). The complement of A is

a set containing all elements of S not in A

( )A

Basic Definitions

Complement of a Set

A

A

S

Venn Diagram illustrating the Complement of an eventVenn Diagram illustrating the Complement of an event

Intersection (And)– a set containing all elements in both A and B

Union (Or)– a set containing all elements in A or B or both

( )A B∩

( )A B∪

Basic Definitions (Continued)

A B∩

Sets: A Intersecting with B

AB

S

Sets: A Union B

A B∪

AB

S

1. A B A B∪ = ∩

2. A B A B∩ = ∪

DeMoivre’s laws

=

=

( ) ( ) A A B A B= ∩ ∪ ∩

Another useful rule

=

In wordsThe event A occurs if A occurs and B occurs or A occurs and B doesn’t occur.

• Mutually exclusive or disjoint sets

–sets having no elements in common, having no intersection, whose intersection is the empty set

• Partition

–a collection of mutually exclusive sets which together include all possible elements, whose union is the universal set

Basic Definitions (Continued)

Mutually Exclusive or Disjoint Sets

A B

S

Sets have nothing in common

Sets: Partition

A1

A2

A3

A4

A5

S

In many examples the sample space S = {o1, o2, o3, … oN} has a finite number, N, of oucomes.

Also each of the outcomes is equally likely (because of symmetry).

Then P[{oi}] = 1/N and for any event E

[ ] ( )( )

( ) no. of outcomes in =

total no. of outcomes

n E n E EP E

n S N= =

( ): = no. of elements of n A ANote

Probability: Classical Approach

[ ] ( )( )

( ) no. of outcomes in =

total no. of outcomes

n E n E EP E

n S N= =

Range of Values for P(A):

Complements - Probability of not A

Intersection - Probability of both A and B

Mutually exclusive events (A and C) :

Range of Values for P(A):

Complements - Probability of not A

Intersection - Probability of both A and B

Mutually exclusive events (A and C) :

1)(0 ≤≤ AP

P A P A( ) ( )= −1

P A B n A Bn S

( ) ( )( )

∩ = ∩

P A C( )∩ =0

Basic Rules for Probability 2-25

E = the event that “sum of rolled point is 7” ={ (6, 1), (5, 2), (4, 3), (3, 4), (3, 5), (1, 6)}

Rolling a two balanced dice.

Example: What is the probability of randomly drawing

either an Ace or a 7 from a deck of 52 playing cards?

•P(Card is an Ace) 4/52

•P(Card is a 7) 4/52

•P(Card is an Ace AND a 7) 0

P(Draw an Ace OR Draw a 7) ?

= P(Ace) + P(7) – P(Ace and 7)

= 4/52 + 4/52 – 0/52) = 8/52

P(A or B) = P(A) + P(B) – P(A and B)

Example: What is the probability of randomly drawing

either an ace or a heart from a deck of 52 playing cards?

•P(Ace) 4/52

•P(Heart) 13/52

•If we simply added them, we would get 17/52. This is NOT the correct result!

•P(Ace and Heart) 1/52

•There is one NON-disjoint event present. Notice how the Ace of Hearts has been counted twice. Therefore we must subtract this doubled item. So the

correct answer is: (4/52 + 13/52 – 1/52) = 16/52.

P(A or B) = P(A) + P(B) – P(A and B)

Pick a Card: Sample Space

Event ‘Ace’Union of Events ‘Heart’and ‘Ace’

Event ‘Heart’

The intersection of theevents ‘Heart’ and ‘Ace’ comprises the single pointcircled twice: the ace of hearts

P Heart Ace

n Heart Ace

n S

( )

( )

( )

=

=

=16

52

4

13

P Heartn Heart

n S

( )( )

( )

= = =13

52

1

4

P Acen Ace

n S

( )( )

( )

= = =4

52

1

13

P Heart Acen Heart Ace

n S

( )( )

( )

= =1

52

Hearts Diamonds Clubs Spades

A A A AK K K KQ Q Q QJ J J J

10 10 10 109 9 9 98 8 8 87 7 7 76 6 6 65 5 5 54 4 4 43 3 3 32 2 2 2

2-29

Let's roll a die once.

S = {1, 2, 3, 4, 5, 6}This is the sample space---all the possible outcomes

( ) Number of ways that can occur

Number of possibilities

EP E =

probability an event will occur

What is the probability you will roll an even number?

There are 3 ways to get an even number, rolling a 2, 4 or 6

There are 6 different numbers on the die.

( ) 3 1Even number

6 2P = =

http://budgetstockphoto.com/samples/pics/dice.jpg

The word and in probability means the intersection of two events.

What is the probability that you roll an even number and a number greater than 3?

E = rolling an even number F = rolling a number greater than 3

( )P E F∩How can E occur? {2, 4, 6}

How can F occur? {4, 5, 6}

{2,4,6} {4,5,6} {4,6}E F∩ = ∩ =

2 1

6 3= =

The word or in probability means the union of two events.

What is the probability that you roll an even number or a number greater than 3?

( )P E F∪ 4 2

6 3= = {2, 4,6} {4,5,6} {2,4,5,6}E F∪ = ∪ =

Another Example:We are shooting at an archery target with radius R. The bullseye has radius R/4. There are three other rings with width R/4. We shoot at the target until it is hit

R

S = set of all points in the target

= {(x,y)| x2 + y2 ≤ R2}

E, any event is a sub region (subset) of S.

E, any event is a sub region (subset) of S.

E

[ ] ( )( )

( )2

: =Area E Area E

P EArea S Rπ

=Define

[ ]

2

2

1416

R

P BullseyeR

π

π

÷ = =

[ ]

2 2

2

3 29 4 54 4

16 16

R RP White ring

R

π π

π

− ÷ ÷ − = = =

Probability – Basic ConceptsOdds

A comparison of the number of favorable outcomes to the number of unfavorable outcomes.

Odds are used mainly in horse racing, dog racing, lotteries and other gambling games/events.

Odds in Favor: number of favorable outcomes (A) to the number of unfavorable outcomes (B).

Example:

A to B A : B

What are the odds in favor of rolling a 2 on a fair six-sided die?

1 : 5

What is the probability of rolling a 2 on a fair six-sided die?1/6


Odds against: number of unfavorable outcomes (B) to the number of favorable outcomes (A).

Example:

What are the odds against rolling a 2 on a fair six-sided die?

B to A B : A

5 : 1 What is the probability against rolling a 2 on a fair six-sided die?

5/6


Two hundred tickets were sold for a drawing to win a new television. If you purchased 10 tickets, what are the odds in favor of you winning the television?

Example:

200 – 10 =

10 : 190 What is the probability of winning the television?

10/200

190 Unfavorable outcomes

10 Favorable outcomes

= 1 : 19

1/20= = 0.05

Probability – Basic ConceptsConverting Probability to Odds

The probability of rain today is 0.43. What are the odds of rain today?

Example:

100 – 43 =

43 : 57 The odds for rain today:

57Unfavorable outcomes:

P(rain) = 0.43

Of the 100 total outcomes, 43 are favorable for rain.

57 : 43 The odds against rain today:

Probability – Basic ConceptsConverting Odds to Probability

The odds of completing a college English course are 16 to 9. What is the probability that a student will complete the course?

Example:

16 : 9 The odds for completing the course:

P(completing the course) =

Favorable outcomes + unfavorable outcomes = total outcomes 16 + 9 = 25

= 0.64

• Conditional Probability - Probability of A given B

Independent events:

• Conditional Probability - Probability of A given B

Independent events:

0)( ,)(

)()( ≠∩= BPwhereBP

BAPBAP

P A B P A

P B A P B

( ) ( )

( ) ( )

==

2-4 Conditional Probability2-40

Rules of conditional probability:Rules of conditional probability:

If events A and D are statistically independent:

so

so

P A B P A BP B

( ) ( )( )

= ∩ P A B P A B P B

P B A P A

( ) ( ) ( )

( ) ( )

∩ ==

P AD P A

P D A P D

( ) ( )

( ) ( )

=

=)()()( DPAPDAP =∩

Conditional Probability (continued)

2-41

P A B P A

P B A P B

and

P A B P A P B

( ) ( )

( ) ( )

( ) ( ) ( )

==

=

Conditions for the statistical independence of events A and B:

P Ace HeartP Ace Heart

P Heart

P Ace

( )( )

( )

( )

=

= = =

1521352

113 P(Heart)===

P(Ace)

e)P(HeartIAc=Ace)|P(Heart

4

1

52

4521

)()(52

1

52

13*

52

4)( HeartPAcePHeartAceP ===

Independence of Events2-42

The probability of the union of several independent events is 1 minus the product of probabilities of their complements:

P A A A An P A P A P A P An( ) ( ) ( ) ( ) ( )1 2 3

11 2 3

∪ ∪ ∪ ∪ = −

The probability of the intersection of several independent events is the product of their separate individual probabilities:

P A A A An P A P A P A P An( ) ( ) ( ) ( ) ( )1 2 3 1 2 3

∩ ∩ ∩ ∩ =

Product Rules for Independent Events

2-43

P A P A B P A B( ) ( ) ( )= ∩ + ∩

In terms of conditional probabilities:

More generally (where Bi make up a partition):

P A P A B P A BP A B P B P A B P B

( ) ( ) ( )( ) ( ) ( ) ( )

= ∩ + ∩= +

P A P A Bi

P ABi

P Bi

( ) ( )

( ) ( )

= ∩∑= ∑

The Law of Total Probability and Bayes’ Theorem

The law of total probability:

2-44

Event U: Stock market will go up in the next yearEvent W: Economy will do well in the next year

.66.06.60

.20.30.80.75

.2.81.80

30.

.75

=+=

))((+))((=

)W)P(W|P(U+W)P(W)|P(U=

)WP(U+W)P(U=P(U)

==)WP(=P(W)

=)W|P(U

=W)|P(U

∩∩

−⇒

The Law of Total Probability

2-45

• Bayes’ theorem enables you, knowing just a little more than the probability of A given B, to find the probability of B given A.

• Based on the definition of conditional probability and the law of total probability.

P B AP A B

P A

P A B

P A B P A B

P AB P B

P AB P B P AB P B

( )( )

( )

( )

( ) ( )

( ) ( )

( ) ( ) ( ) ( )

=

=+

=+

Applying the law of total probability to the denominator

Applying the definition of conditional probability throughout

Bayes’ Theorem2-46

• Given a partition of events B1,B2 ,...,Bn:

P B AP A B

P A

P A B

P A B

P A B P B

P A B P B

i

i i

( )( )

( )

( )

( )

( ) ( )

( ) ( )

1

1

1

1 1

= ∩

= ∩∩∑

=∑

Applying the law of total probability to the denominator

Applying the definition of conditional probability throughout

Bayes’ Theorem Extended2-47

An economist believes that during periods of high economic growth, the U.S. dollar appreciates with probability 0.70; in periods of moderate economic growth, the dollar appreciates with probability 0.40; and during periods of low economic growth, the dollar appreciates with probability 0.20.

During any period of time, the probability of high economic growth is 0.30, the probability of moderate economic growth is 0.50, and the probability of low economic growth is 0.50.

Suppose the dollar has been appreciating during the present period. What is the probability we are experiencing a period of high economic growth?

Partition:H - High growth P(H) = 0.30M - Moderate growth P(M) = 0.50L - Low growth P(L) = 0.20

Event A Appreciation−==

=

P A HP A MP A L

( ) .( ) .( ) .

0 700 40

0 20

Baye’s Theorem Extended

2-48

P H AP H A

P AP H A

P H A P M A P L AP A H P H

P A H P H P A M P M P A L P L

( )( )

( )( )

( ) ( ) ( )( ) ( )

( ) ( ) ( ) ( ) ( ) ( )( . )( . )

( . )( . ) ( . )( . ) ( . )( . ).

. . ...

.

=

=+ +

=+ +

=+ +

=+ +

=

=

0 70 0 300 70 0 30 0 40 050 0 20 0 20

0 210 21 0 20 0 04

0 210 45

0 467

Example (continued)2-49

Example Three jars contain colored balls as described in

the table below. One jar is chosen at random and a ball is selected. If the

ball is red, what is the probability that it came from the 2nd jar?

Jar # Red White Blue

1 3 4 1

2 1 2 3

3 4 3 2

Example We will define the following events:

J1 is the event that first jar is chosen

J2 is the event that second jar is chosen

J3 is the event that third jar is chosen

R is the event that a red ball is selected

Example The events J1 , J2 , and J3 mutually exclusive

Why? You can’t chose two different jars at the same time

Because of this, our sample space has been divided or partitioned along these three events

Venn Diagram Let’s look at the Venn Diagram

Venn Diagram All of the red balls are in the first, second, and

third jar so their set overlaps all three sets of our partition

Finding Probabilities What are the probabilities for each of the

events in our sample space? How do we find them?

( ) ( ) ( )BPBAPBAP |=∩

Computing Probabilities

Similar calculations show:

( ) ( ) ( )8

1

3

1

8

3| 111 =⋅==∩ JPJRPRJP

( ) ( ) ( )

( ) ( ) ( )27

4

3

1

9

4|

18

1

3

1

6

1|

333

222

=⋅==∩

=⋅==∩

JPJRPRJP

JPJRPRJP

Venn Diagram Updating our Venn Diagram with these

probabilities:

Where are we going with this? Our original problem was:

One jar is chosen at random and a ball is selected. If the ball is red, what is the probability that it came from the 2nd jar?

In terms of the events we’ve defined we want:

( ) ( )( )RP

RJPRJP

∩= 22 |

Finding our Probability

( ) ( )( )

( )( ) ( ) ( )RJPRJPRJP

RJP

RP

RJPRJP

∩+∩+∩∩=

∩=

321

2

22 |

We already know what the numerator portion is from our Venn Diagram

What is the denominator portion?

Arithmetic! Plugging in the appropriate values:

( ) ( )( ) ( ) ( )

17.071

12

274

181

81

181

|321

22

≈=

+

+

=

∩+∩+∩∩=

RJPRJPRJP

RJPRJP

Thanking You

?

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Dr Ashok Kumar Probability